According to an alm anac 70% of adult smokers started smoking before turing 18 y

Question

According to an alm anac 70% of adult smokers started smoking before turing 18 years old when technology is used use the Tech He button for further assistance (a) Compute the mean and standard deviation of the random variable X, the number of smokers who started before 18 in 200 trials of the probability experiment (b) Interpret the mean (e) Would it be unusual to observe 160 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers? Why? [] (Round to the nearest tenth as needed.) b) What is the correct interpretation of the mean? 0 A. It is expected that in 50% of candom samples of 200 adult smokers 140 will have started smoking before turning 18 O B· is expected that in a random sample of 200 adult smokers. 140 will have started smoking before turning 18. C. It is expected that in a random sample of 200 adult smokers, 140 will have started smoking after turning 18 ic) Would it be unusual to observe 160 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers? 0 A. No because 160 is less than-2n B. No because 160 is greater than 2n C. Yes because 160 is greater than + 2 D. No because 160 s between g-2n and 2 E. Yes because 160 is between u-2a and + 2n. Phok to seleot your arc Type here to search Esc FT F2 F3 F4 FS F6 F7 F8 F10 F11

Explanation / Answer

Answer to all questions below with detailed explanations and calculations:

p = .7

a.

n = 200
So, Mean = np = .7*200 = 140
Stdev = sqrt(npq) = sqrt(.7*.3*200) = 6.481

b. C is wrong. It is not 'after' its 'before'. A is also wrong interpretation.

B is the correct answer

c. Lets calculate Z test statistic first. 160 is 140 + more than 2 deviation away from mean. (140+2*6.481)
Something with more than deviations from the mean is unusual. Hence, right answer is C
C says that 160 is greater than Mean +2sigma, hence its the right answer

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